In a hydrogen atom, the binding energyo of the electron in the gound state is `E_(1)`. Then the frequecny of revolution of the electron in the nth orbit is
A. `(2E_(1))/(n^(3)h)`
B. `(2E_(1)n^(3))/(h)`
C. `sqrt((2mE_(1))/(n^(3)h))`
D. `(E_(1)n^(2))/(h)`

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1 Answers

Correct Answer - A
`E_(1)=(me^(4))/(8epsi_(0)^(2)h^(2))`
`f_(n)=(v_(n))/(2pir_(n))=((e^(2)//2epsi_(0)nh))/((2pi)epsi_(0)n^(2)h^(2)//pime^(2))=(me^(2))/(4epsi_(0)^(2)n^(3)h^(3))=(8E_(1)epsi_(0)^(2)h^(2))/(4epsi_(0)^(2)n^(3)h^(3))=(2E_(1))/(n^(3)h)`

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